$e.q.\Rightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=2 \cdot 2003, (x+y+z),(x-y)^2+(y-z)^2+(z-x)^2 \in \mathbb{N}$ $case(i):(x+y+z)=2, (x-y)^2+(y-z)^2+(z-x)^2=2003 \Rightarrow$ no solution $case(ii):(x+y+z)=2003 \cdots (1), (x-y)^2+(y-z)^2+(z-x)^2=2 \cdots (2)$ $W.L.O.G$, Let $x \le y \le z$ $(2)\Rightarrow (x-y,y-z,z-x)=(-1,0,1)$ so (x,y,z)=(667,668,668) Answes:(x,y,z)=(667,668,668),(668,667,668),(668,668,667)

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